# Solution to the Two Circles Inside Two Polygons Inside a Circle Puzzle +-- {.image} [[TwoCirclesInsideTwoPolygonsInsideaCircle.png:pic]] > Two circles, inside two regular polygons, inside a large circle. What fraction is shaded? =-- ## Solution by [[Lengths in Equilateral Triangles]], [[Properties of Chords]], and [[Intersecting Chords Theorem]] +-- {.image} [[TwoCirclesInTwoPolysInaCircleLabelled.png:pic]] =-- As the line segment $E F$ is a [[chord]], its [[perpendicular bisector]] $A D$ passes through the centre of the circle and so is a diameter. Let $x$ be the length of half the side of the square, so $x$ is the length of $E C$. Using [[lengths in an equilateral triangle]], the length of $A B$ is $\sqrt{3} x$ so the length of $A C$ is $2 x + \sqrt{3} x$. Write $y$ for the length of $C D$, then using the [[intersecting chords theorem]] applied to $E F$ and $A D$, the $x$ and $y$ satisfy: $$ (2 x + \sqrt{3} x) y = x^2 $$ which rearranges to $$ y = \frac{x}{2 + \sqrt{3}} $$ The diameter of the outer circle is therefore: $$ 2 x + \sqrt{3} x + y = 2x + \sqrt{3} x + \frac{x}{2 + \sqrt{3}} = \frac{(8 + 4 \sqrt{3}) x}{2 + \sqrt{3}} = 4 x $$ Its area is therefore $4 \pi x^2$. The radius of the circle in the square is $x$, and the radius of the circle in the equilateral triangle is a third of the length of $A B$, so is $\frac{1}{\sqrt{3}} x$. The area of the shaded regions is therefore $\pi x^2 + \frac{1}{3} \pi x^2 = \frac{4}{3} \pi x^2$. Therefore $\frac{1}{3}$rd of the outer circle is shaded.